Permutations and Combinations 2 Question 22
22. 7 relatives of a man comprises 4 ladies and 3 gentlemen, his wife has also 7 relatives ; 3 of them are ladies and 4 gentlemen. In how many ways can they invite a dinner party of 3 ladies and 3 gentlemen so that there are 3 of man’s relative and 3 of the wife’s relatives?
$(1985,5$ M)
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Answer:
Correct Answer: 22. (485)
Solution:
- The possible cases are
Case I A man invites 3 ladies and women invites 3 gentlemen.
Number of ways $={ }^{4} C _3 \cdot{ }^{4} C _3=16$
Case II A man invites (2 ladies, 1 gentleman) and women invites (2 gentlemen, 1 lady).
Number of ways $=\left({ }^{4} C _2 \cdot{ }^{3} C _1\right) \cdot\left({ }^{3} C _1 \cdot{ }^{4} C _2\right)=324$
Case III A man invites (1 lady, 2 gentlemen) and women invites (2 ladies, 1 gentleman).
Number of ways $=\left({ }^{4} C _1 \cdot{ }^{3} C _2\right) \cdot\left({ }^{3} C _2 \cdot{ }^{4} C _1\right)=144$
Case IV A man invites (3 gentlemen) and women invites (3 ladies).
Number of ways $={ }^{3} C _3 \cdot{ }^{3} C _3=1$
$\therefore$ Total number of ways,
$$ =16+324+144+1=485 $$
